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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2014
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A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is \({\theta _{iC}}\) and Brewster's angle of incidence is \({\theta _{iB}}\), such that \(\sin \,{\theta _{iC}}/\sin \,{\theta _{iB}} = \eta  = 1.28\). The  relative refractive index of the two media is

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Answer & Solution

Correct Answer

(C) \(0.8\)

Step-by-step Solution

Detailed explanation

Here, \(\sin \theta_{i c} / \sin \theta_{i B}=1.28\)

As we know,

\(\mu=\frac{\sin \theta_{\mathrm{iB}}}{\sin \left(\frac{\pi}{2}-\theta_{\mathrm{iB}}\right)}\)

where, \(\theta_{\mathrm{iB}}\) is Brewster's angle of incidence,

And, \(\mu=\frac{1}{\sin \theta_{\text {ic }}}\)

Onsolving we get, relative refractive index of the two media.

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